Question: Let m[O)...m[4] be mutexes (binary semaphores) and P[O)...P[4] be processes. Suppose each process P[i] executes the following: wait(m[i]); wait(m[(i+1) mod 4]); release(m[i]); release(m[(i+1) mod 4);

 Let m[O)...m[4] be mutexes (binary semaphores) and P[O)...P[4] be processes. Suppose

each process P[i] executes the following: wait(m[i]); wait(m[(i+1) mod 4]); release(m[i]); release(m[(i+1)

Let m[O)...m[4] be mutexes (binary semaphores) and P[O)...P[4] be processes. Suppose each process P[i] executes the following: wait(m[i]); wait(m[(i+1) mod 4]); release(m[i]); release(m[(i+1) mod 4); What is the effect of putting the mod 4 operation? Select one: O A. It will create the effect of a loop. OB. It will keep the value of the semaphore to be less or equal to 4 OC. It ensured that only the 5 semaphores will be accessed. OD. It will increase the value of 1(i=i+1); What is the effect of the cobegin and coned statement in the following code? var a,b,c,d,e,f,g: semaphore; begin cobegin begin si; V(a); V(b) end; begin P(a); 52; V(c); V(d) end; begin P(b); $3; V(e) end; begin P(c); P(E); s4 end; begin P(d); P(); P(g): 55 end begin $6; V(): V(g) end; coend ond; Select one: O A It allows for all concurrent execution of the statements 51, 52 and 53. O B. It ensures mutual exclusion is imposed. OC It ensures that the statements in between them is executed in order OD. It functions like a loop which executes statements 51.52 and 53 forever

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